Power law for indices

﻿Hello, my name is Mr. Clasper. And today we're going to learn how to combine index laws. If you haven't done so already, I would strongly recommend that you look at our lessons on the multiplication law for indices, the division law for indices, and the power law for indices. As we're going to be using all three of these in today's lesson. Let's take a look at this example, x to the power of two, all raised to the power of four, then multiplied by x to the power of negative seven. If we focus on this part first, we can apply our power law for indices. So when we apply the power law, it means that we can multiply our powers and keep the base the same. So this will be equivalent to x to the power of eight, giving us a new expression of x to the power of eight multiplied by x to the power of negative seven. Next we can apply our multiplication law. So our multiplication law states that when bases are the same, we can add our powers together. So when we add eight and negative seven, we get a value of one. So we get x to the power of one. This can also be written as x. Let's try this example. For this example, we're going to simplify the numerator first. So using our multiplication law, we need to multiply our coefficients, which would give us 30, and we need to add our powers of x. So this would give us 16. So our new numerator would become 30x to the power of 16. And we're still going to divide this by 15x to the power of 11. Now we can use our division law. So our division law says that we can divide our coefficients and we can subtract our powers of x. So this would leave us with 2x to the power of five. This is our last example. We're going to simplify the numerator and the denominator. Both of these can be simplified by using the multiplication law. So using the multiplication law, we can multiply our coefficients, and we can add our powers. So the numerator would become 30x to the power of 16. And the denominator is equivalent to 45x to the power of 19. At this point, all that remains is for us to use our division law. So we divide our coefficients, and we subtract our powers, which in this example would give us two over 3x to the power of negative three. If you're wondering where two over three has come from, think about 30 over 45 as a fraction, and then simplify it. So if we simplify 30 over 45, we get two over three. And our power of negative three came from calculating 16, subtract 19. Here are some questions for you to try, pause the video, to complete your task, and click resume once you're finished. And here are the answers. Let's take a look at question 1d, so we have an answer of a to the power of negative four. If we look at the numerator, when we simplify this, we would get a to the power of nine, and the denominator would simplify to a to the power of 13. And if we divide a to the power of nine by a to the power of 13, we get a to the power of negative four. Looking at question two, the true or false statements. So part a was false. So they should be a to the power of 10. This is because a to the power of two, or the power of three would simplify to a to the power of six. And that if we multiply this by a to the power of four, that should give us a to the power of 10. So with the multiplication law, we add our powers. And for part 2d this is also a false, we should get 24t to the power of 16. So once again, one of the key mistakes that's been made here is that the coefficient two should be raised to the power of three. So when we simplify our bracket, this would give us 8t to the power of 15. And then if we multiply that by 3t, this would give us 24t to the power of 16. Here are some questions for you to try, pause the video, to complete your task, and click resume once you're finished. And here are your answers. Let's take a look at 3a. So if we know that m is three, if we simplify the denominator, we would get a, to the power of five. And then if we calculate a, to the power of nine divided by a to the power of five, this does give us a to the power of four, therefore backing up the fact that we have m is equal to three as a solution for this problem. If you look at part b, if we try to find m, first of all, we know that we need to multiply three and m together. Then we need to add two to this, to get a value of 17, or a new power of 17. This means that 3m must be worth 15, as 15 plus two is 17. Now that we know that 3m is worth 15, we can also work out that m must be five. So we have a value for m. To find our coefficient, this is a little bit trickier. So we have, we need to raise two to the power of three, first of all. So two to the power of three, will give us a value of eight. Then we need to multiply eight by p to get our new coefficient of 24. So this means that p would have to be three. And for question four. If we went backwards to the function machine, using inverse operations, we can find our solution. So 36a to the power of four, divided by six, then multiplied by 2a to the power of three, then divided by 4a to the power of five, would give us our solution of 3a to the power of two. And that brings us to the end of our lesson. So at this point, you should be confident by applying your multiplication law, your division law, and your power law for indices, as well as combining them. Why not have a go at the exit quiz before you leave? Just to see how well you've done. Hopefully I'll see you soon.

Combining index laws

In this lesson, we will be applying all three Index Laws to help us simplify more complicated expressions.

Lesson overview:Combining Index Laws

Core Content

Rules of Indices: